Lecture 21

class: center, middle, inverse, title-slide

.title[
# Lecture 21
]
.subtitle[
## Multiple Linear Regression
]
.author[
### Psych 10 C
]
.institute[
### University of California, Irvine
]
.date[
### 05/18/2022
]

---

## Linear regression

- Last class we finished our example with multiple linear regression in a mental
rotation task.

- Today we will look at another example, however, this time we will not have a 
hypothesis, so we will have to take a "brute force" approach.

- We are interested in studying the effects of age and height on blood 
pressure.

- We are not sure if only one or both of these variables are good predictors; so 
we want to compare all the models we can build using these two variables.

---

## Data

- Now that we have a research question (are height and age good predictors of 
blood pressure), we need to look at the data in the study.

- There are 50 participants in this study, all of whom had their blood pressure 
taken during a routine check up. The average blood pressure of participants was
116.42 (mmHg), with a range
92 to 144.

- The age of the participants ranged from 20 to 
70, with an average of 44.64 years.

- The height of participants ranged from 58.3 to 
75.8 with an average of 66.894 inches.

- From those participants in the study 25 are female and the rest are male.

---

## Data

- Now that we have a description of the data we can visualize our observations
using a scatter plot, in this case we are interested in two variables (age and
height) so we can make two independent graphs.

.pull-left[
<img src="data:image/png;base64,#lec-21_files/figure-html/blood-age-1.png" style="display: block; margin: auto;" />
]

.pull-right[
<img src="data:image/png;base64,#lec-21_files/figure-html/blood-height-1.png" style="display: block; margin: auto;" />
]

---

## Data

- From the previous graphs it seems that both height and age could be associated
with the blood pressure of participants.

- However, we can't draw conclusions from a plot, we need to test linear models
in order to tell if our independent variables are good predictors.

- Given that we have 2 independent variables (without taking into account the
sex of participants which is categorical), we can compare 4 models.

- When we **only** have continuous variables as predictors we don't model 
interactions because there is no way to interpret them.

---

## Models

- The 4 models that we need to compare are:

1. **Null model**: Blood pressure is constant regardless of age and height of 
  a participant `$$\text{blood-pressure}_i \sim \text{Normal}(\beta_0, \sigma^2_1)$$`
  
--

1. **Age model**: Blood pressure is a linear function of the age of the 
  participant `$$\text{blood-pressure}_i \sim \text{Normal}(\beta_0 + \beta_1 \text{age}_i, \sigma^2_2)$$`
  
--

1. **Height model**: Blood pressure is a linear function of the height of the 
  participant `$$\text{blood-pressure}_i \sim \text{Normal}(\beta_0 + \beta_2 \text{height}_i, \sigma^2_3)$$`
--

1. **Age & Height model**: Blood pressure is a linear function of the height of the 
  participant `$$\text{blood-pressure}_i \sim \text{Normal}(\beta_0 + \beta_1 \text{age}_i + \beta_2 \text{height}_i, \sigma^2_4)$$`

---

## Predictions and errors

- As we have done before, we want to calculate the predictions and errors of 
each model in order to make a comparison and select the best one.

- Once we have the model that accounts for the data better, we will look at 
the distribution of the difference between observations and model's predictions
`$(\hat{\epsilon}_i)$` to evaluate the adequacy of the model.

---

## Null model

.pull-left[

```r
# Total sample size
n_total <- nrow(pressure)
# Prediction of the null model
null <- pressure %>% 
 summarise("pred" = mean(blood_pressure)) %>% 
 pull(pred)
# Adding prediction and error of null to the data
pressure <- pressure %>% 
 mutate("prediction_null" = null,
 "error_null" = (blood_pressure - prediction_null)^2)
# Calculating SSE of the null model
sse_null <- sum(pressure$error_null)
# Calculate Mean SE of the null model
mse_null <- 1/n_total * sse_null
# Calculate the BIC of the null model
bic_null <- n_total * log(mse_null) + 1 * log(n_total)
```
]

.pull-right[
 
 
 
<img src="data:image/png;base64,#lec-21_files/figure-html/unnamed-chunk-1-1.png" style="display: block; margin: auto;" />
]

- The estimate of the intercept was `$\hat{\beta}_0$` = 116.42
---

## Age model

.pull-left[

```r
# Get estimators for beta0 and beta1
betas_age <- lm(formula = blood_pressure ~ age, data = pressure)$coef
# Adding prediction and error for the age lm to the data
pressure <- pressure %>% 
 mutate("prediction_age" = betas_age[1] + betas_age[2] * age,
 "error_age" = (blood_pressure - prediction_age)^2)
# Calculating SSE for the age lm
sse_age <- sum(pressure$error_age)
# Calculate Mean SE for the age lm
mse_age <- 1/n_total * sse_age
# Calculate the value of R^2 for the age lm
r2_age <- (sse_null - sse_age) / sse_null
# Calculate the BIC for the age lm
bic_age <- n_total * log(mse_age) + 2 * log(n_total)
```
]

.pull-right[
 
 
 
<img src="data:image/png;base64,#lec-21_files/figure-html/unnamed-chunk-2-1.png" style="display: block; margin: auto;" />
]

- The estimate of the intercept was `$\hat{\beta}_0$` = 91.7,
the estimate of the slope associated to age was `$\hat{\beta}_1$` = 0.55.

---

## Height model